$X,Y,Z$ are disjoint sets of vectors in a Banach space M. $S_X,S_Y,S_Z$ denotes the subspace formed by the vectors in $X, Y$ and $Z$ respectively. $S_{YZ}$ and $S_{XZ}$ denote the subspaces formed by $Y\cup Z$ and $X\cup Z$ respectively. It is given that $\displaystyle \inf_{z\in S_Z}\|x-z\| = \inf_{z\in S_{YZ}}\|x-z\|, \forall x \in X$.
Does it imply that $\displaystyle \inf_{z\in S_Z}\|y-z\| = \inf_{z\in S_{XZ}}\|y-z\|, \forall y \in Y$? (Assume that solution to all minimization problems exist).
Please provide a counter example if this result does not hold for a general Banach space. Also a discussion on conditions on M under which this result holds would be helpful.
$\textbf{Note:}$ It is easy to show that this result holds in a Hilbert space using orthogonal projections and inner products. Here is how it goes for Hilbert spaces.
$\hat{x}_{_Z}$ and $\hat{x}_{_{YZ}}$ be the projection of $x$ on $S_Z$ and $S_{YZ}$ respectively. Let $\hat{y}_{_Z}$ and $\hat{y}_{_{XZ}}$ be the projection of $y$ on $S_Z$ and $S_{XZ}$ respectively.
$\textbf{To prove:}$ $\|y - \hat{y}_{_Z}\| = \|y - \hat{y}_{_{XZ}}\|, \forall y \in Y$
It is given that $\|x - \hat{x}_{_Z}\| = \|x - \hat{x}_{_{YZ}}\|$. By the uniqueness of orthogonal projections, $\hat{x}_{_Z} = \hat{x}_{_{YZ}}$. Now, $\langle x - \hat{x}_{_{YZ}},y\rangle = 0, \forall y \in Y$. Thus, $\langle x - \hat{x}_{_Z},y\rangle = 0, \forall y \in S_Y$. Define $\tilde{x}_{_Z} = x - \hat{x}_{_Z}$, so $\langle y,\tilde{x}_{_Z} \rangle = 0$.
If it is shown that $\langle y - \hat{y}_{_Z},x\rangle = 0, \forall x \in S_X$ then the proof is complete because it is known that $\langle y - \hat{y}_{_{XZ}},x\rangle = 0, \forall x \in S_X$ and by uniqueness of orthogonal projections $\hat{y}_{_Z} = \hat{y}_{_{XZ}}$, thereby implying $\|y - \hat{y}_{_Z}\| = \|y - \hat{y}_{_{XZ}}\|, \forall y \in Y$.
Consider, $\langle y - \hat{y}_{_Z},x\rangle = \langle y - \hat{y}_{_Z},\hat{x}_{_Z} + \tilde{x}_{_Z}\rangle = \langle y - \hat{y}_{_Z},\tilde{x}_{_Z}\rangle = \langle y ,\tilde{x}_{_Z}\rangle = 0$, which completes the proof.